Inverzi hiperboličkih funkcija su površinske hiperboličke funkcije . Naziv dolatzi iz činjenice da one izračunavaju površinu sektor jedinične hiperbole
x
2
−
y
2
=
1
{\displaystyle x^{2}-y^{2}=1}
inverzne trigonometrijske funkcije izračunavaju dužinu luka sektora jediničnog kruga
x
2
+
y
2
=
1.
{\displaystyle x^{2}+y^{2}=1.}
arsinh, arcsinh (u USA-a) ili asinh (u računarstvu). Oznake sinh-1 (x), cosh-1 (x) itd. se, također, koriste, uprkos činjenici da se mora voditi računa kako bi se izbjeglo pogrešno shvatanje eksponenta -1 kao stepena, a ne kao skrećene oznake za inverznu funkciju. Skraćenice arcsinh , arccosh itd. se najčešće koriste, iako to nije pravilno, pošto je prefiks arc skraćenica za arcus , dok prefiks ar označava površinu .[ 1]
funkcija artanh . Vrijednosi inverznih hiperboličkih funkcija su hiperbolički uglovi .
Logaritamsko predstavljanje
uredi
Operatori su definisani u kompleksnoj ravni kao:
arsinh
x
=
ln
(
x
+
x
2
+
1
)
,
arcosh
x
=
ln
(
x
+
x
−
1
x
+
1
)
,
artanh
x
=
ln
(
1
−
x
2
1
−
x
)
=
1
2
ln
(
1
+
x
1
−
x
)
,
arcsch
x
=
ln
(
1
+
1
x
2
+
1
x
)
,
arsech
x
=
ln
(
1
x
−
1
1
x
+
1
+
1
x
)
,
arcoth
x
=
1
2
ln
x
+
1
x
−
1
.
{\displaystyle {\begin{aligned}\operatorname {arsinh} \,x&=\ln(x+{\sqrt {x^{2}+1}}),\\[2.5ex]\operatorname {arcosh} \,x&=\ln(x+{\sqrt {x-1}}{\sqrt {x+1}}),\\[1.5ex]\operatorname {artanh} \,x&=\ln \left({\frac {\sqrt {1-x^{2}}}{1-x}}\right)={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right),\\\operatorname {arcsch} \,x&=\ln \left({\sqrt {1+{\frac {1}{x^{2}}}}}+{\frac {1}{x}}\right),\\\operatorname {arsech} \,x&=\ln \left({\sqrt {{\frac {1}{x}}-1}}{\sqrt {{\frac {1}{x}}+1}}+{\frac {1}{x}}\right),\\\operatorname {arcoth} \,x&={\frac {1}{2}}\ln {\frac {x+1}{x-1}}.\end{aligned}}}
gornji kvadratni korijeni su osnovni kvadratni korijeni . Za realne argumente koji daje realnu vrijednost, mogu se naći određena pojednostavljenja, npr.
x
−
1
x
+
1
=
x
2
−
1
{\displaystyle {\sqrt {x-1}}{\sqrt {x+1}}={\sqrt {x^{2}-1}}}
Inverzne hiperboličke funkciej u kompleksnoj ravni
arsinh
(
z
)
{\displaystyle \operatorname {arsinh} (z)}
arcosh
(
z
)
{\displaystyle \operatorname {arcosh} (z)}
artanh
(
z
)
{\displaystyle \operatorname {artanh} (z)}
arcoth
(
z
)
{\displaystyle \operatorname {arcoth} (z)}
arsech
(
z
)
{\displaystyle \operatorname {arsech} (z)}
arcsch
(
z
)
{\displaystyle \operatorname {arcsch} (z)}
uredi
Za gornje funkcije mogu se dobiti njihova proširenja u redove:
arsinh
x
{\displaystyle \operatorname {arsinh} \,x}
=
x
−
(
1
2
)
x
3
3
+
(
1
⋅
3
2
⋅
4
)
x
5
5
−
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
7
7
+
⋯
{\displaystyle =x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots }
=
∑
n
=
0
∞
(
(
−
1
)
n
(
2
n
)
!
2
2
n
(
n
!
)
2
)
x
2
n
+
1
(
2
n
+
1
)
,
|
x
|
<
1
{\displaystyle =\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1}
arcosh
x
{\displaystyle \operatorname {arcosh} \,x}
=
ln
2
x
−
(
(
1
2
)
x
−
2
2
+
(
1
⋅
3
2
⋅
4
)
x
−
4
4
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
−
6
6
+
⋯
)
{\displaystyle =\ln 2x-\left(\left({\frac {1}{2}}\right){\frac {x^{-2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-6}}{6}}+\cdots \right)}
=
ln
2
x
−
∑
n
=
1
∞
(
(
−
1
)
n
(
2
n
)
!
2
2
n
(
n
!
)
2
)
x
−
2
n
(
2
n
)
,
x
>
1
{\displaystyle =\ln 2x-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{(2n)}},\qquad x>1}
artanh
x
=
x
+
x
3
3
+
x
5
5
+
x
7
7
+
⋯
=
∑
n
=
0
∞
x
2
n
+
1
(
2
n
+
1
)
,
|
x
|
<
1
{\displaystyle \operatorname {artanh} \,x=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1}
arcsch
x
=
arsinh
x
−
1
{\displaystyle \operatorname {arcsch} \,x=\operatorname {arsinh} \,x^{-1}}
=
x
−
1
−
(
1
2
)
x
−
3
3
+
(
1
⋅
3
2
⋅
4
)
x
−
5
5
−
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
−
7
7
+
⋯
{\displaystyle =x^{-1}-\left({\frac {1}{2}}\right){\frac {x^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-7}}{7}}+\cdots }
=
∑
n
=
0
∞
(
(
−
1
)
n
(
2
n
)
!
2
2
n
(
n
!
)
2
)
x
−
(
2
n
+
1
)
(
2
n
+
1
)
,
|
x
|
<
1
{\displaystyle =\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|<1}
arsech
x
=
arcosh
x
−
1
{\displaystyle \operatorname {arsech} \,x=\operatorname {arcosh} \,x^{-1}}
=
ln
2
x
−
(
(
1
2
)
x
2
2
+
(
1
⋅
3
2
⋅
4
)
x
4
4
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
6
6
+
⋯
)
{\displaystyle =\ln {\frac {2}{x}}-\left(\left({\frac {1}{2}}\right){\frac {x^{2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{6}}{6}}+\cdots \right)}
=
ln
2
x
−
∑
n
=
1
∞
(
(
−
1
)
n
(
2
n
)
!
2
2
n
(
n
!
)
2
)
x
2
n
2
n
,
0
<
x
≤
1
{\displaystyle =\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1}
arcoth
x
=
artanh
x
−
1
{\displaystyle \operatorname {arcoth} \,x=\operatorname {artanh} \,x^{-1}}
=
x
−
1
+
x
−
3
3
+
x
−
5
5
+
x
−
7
7
+
⋯
{\displaystyle =x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots }
=
∑
n
=
0
∞
x
−
(
2
n
+
1
)
(
2
n
+
1
)
,
|
x
|
>
1
{\displaystyle =\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1}
Asimptotsko proširenje za arsinh x je dato sa
arsinh
x
=
ln
2
x
+
∑
n
=
1
∞
(
−
1
)
n
−
1
(
2
n
−
1
)
!
!
2
n
(
2
n
)
!
!
1
x
2
n
{\displaystyle \operatorname {arsinh} \,x=\ln 2x+\sum \limits _{n=1}^{\infty }{\left({-1}\right)^{n-1}{\frac {\left({2n-1}\right)!!}{2n\left({2n}\right)!!}}}{\frac {1}{x^{2n}}}}
d
d
x
arsinh
x
=
1
1
+
x
2
d
d
x
arcosh
x
=
1
x
2
−
1
d
d
x
artanh
x
=
1
1
−
x
2
d
d
x
arcoth
x
=
1
1
−
x
2
d
d
x
arsech
x
=
−
1
x
(
x
+
1
)
1
−
x
1
+
x
d
d
x
arcsch
x
=
−
1
x
2
1
+
1
x
2
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsinh} \,x&{}={\frac {1}{\sqrt {1+x^{2}}}}\\{\frac {d}{dx}}\operatorname {arcosh} \,x&{}={\frac {1}{\sqrt {x^{2}-1}}}\\{\frac {d}{dx}}\operatorname {artanh} \,x&{}={\frac {1}{1-x^{2}}}\\{\frac {d}{dx}}\operatorname {arcoth} \,x&{}={\frac {1}{1-x^{2}}}\\{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {-1}{x(x+1)\,{\sqrt {\frac {1-x}{1+x}}}}}\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {-1}{x^{2}\,{\sqrt {1+{\frac {1}{x^{2}}}}}}}\\\end{aligned}}}
For real x :
d
d
x
arsech
x
=
∓
1
x
1
−
x
2
;
ℜ
{
x
}
≷
0
d
d
x
arcsch
x
=
∓
1
x
1
+
x
2
;
ℜ
{
x
}
≷
0
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {\mp 1}{x\,{\sqrt {1-x^{2}}}}};\qquad \Re \{x\}\gtrless 0\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {\mp 1}{x\,{\sqrt {1+x^{2}}}}};\qquad \Re \{x\}\gtrless 0\end{aligned}}}
Primjer derivacije: Neka je θ = arsinh x , tako da je:
d
arsinh
x
d
x
=
d
θ
d
sinh
θ
=
1
cosh
θ
=
1
1
+
sinh
2
θ
=
1
1
+
x
2
{\displaystyle {\frac {d\,\operatorname {arsinh} \,x}{dx}}={\frac {d\theta }{d\sinh \theta }}={\frac {1}{\cosh \theta }}={\frac {1}{\sqrt {1+\sinh ^{2}\theta }}}={\frac {1}{\sqrt {1+x^{2}}}}}
^ As stated by Jan Gullberg, Mathematics: From the Birth of Numbers (New York: W. W. Norton & Company, 1997), ISBN 0-393-04002-X , pg. 539:
Another form of notation, arcsinh x , arccosh x , etc. , is a practice to be condemned as these functions have nothing whatever to do with arc , but with ar ea, as is demonstrated by their full Latin names,
arsinh area sinus hyperbolicus area cosinus hyperbolicus, etc.
![{\displaystyle {\begin{aligned}\operatorname {arsinh} \,x&=\ln(x+{\sqrt {x^{2}+1}}),\\[2.5ex]\operatorname {arcosh} \,x&=\ln(x+{\sqrt {x-1}}{\sqrt {x+1}}),\\[1.5ex]\operatorname {artanh} \,x&=\ln \left({\frac {\sqrt {1-x^{2}}}{1-x}}\right)={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right),\\\operatorname {arcsch} \,x&=\ln \left({\sqrt {1+{\frac {1}{x^{2}}}}}+{\frac {1}{x}}\right),\\\operatorname {arsech} \,x&=\ln \left({\sqrt {{\frac {1}{x}}-1}}{\sqrt {{\frac {1}{x}}+1}}+{\frac {1}{x}}\right),\\\operatorname {arcoth} \,x&={\frac {1}{2}}\ln {\frac {x+1}{x-1}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/340de129855332a05bbab6d14ef8d70cf8c34e50)
